H=-16t^2+50t+7

Solution for H=-16t^2+50t+7 equation:

=-16H^2+50H+7

We move all terms to the left:

-(-16H^2+50H+7)=0

We get rid of parentheses

16H^2-50H-7=0

a = 16; b = -50; c = -7;
Δ = b2-4ac
Δ = -502-4·16·(-7)
Δ = 2948
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2948}=\sqrt{4*737}=\sqrt{4}*\sqrt{737}=2\sqrt{737}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-2\sqrt{737}}{2*16}=\frac{50-2\sqrt{737}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+2\sqrt{737}}{2*16}=\frac{50+2\sqrt{737}}{32}
$